What average force is needed to accelerate a 9.20-gram pellet from rest to 125 m/s over a distance of 0.800 m along the barrel of a rifle?

Given:

u = 0, initial velocity

v = 125 m/s, final velocity

s = 0.0800 m, distance traveled

9.20 g, the mass of the pellet

If the acceleration is a, then

0² + 2 * (a m/s²) * (0.800 m) = (125 m/s) ²

1.6a = 15625

a = 9765.625 m/s²

Calculate the force.

F = (9.20 x 10⁻³ kg) * (9765.625 m/s²) = 98.84 N

Answer: 98.8 N (nearest tenth)